Q. 58 - Calculus

Question

Find the Maclaurin series for the functions in Exercises 51–60

by substituting into a known Maclaurin series. Also, give the

interval of convergence for the series.

$\frac{e^x-e^{-x}}{2}$

Step-by-Step Solution

Answer

The answer is $\frac{e^x-e^{-x}}{2}=\sum _{k=0}^{\infty }\frac{1}{(2k+1)!}{x}^{2k+1}$

Step 1. Given Information

Consider the function $\frac{e^x-e^{-x}}{2}$

Step 2: Find the interval of convergence for the series.

We know that the Maclaurin series for the func$e^x=\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$

So, the series for $h\left(x\right)=e^{-x}$ can be found by substituting $x$ by $-x$

That is, $e^{-x}=\sum _{k=0}^{\infty }\frac{1}{k!}{(-x)}^k$

Finally, to find the series for the function $f\left(x\right)=\frac{e^x-e^{-x}}{2}$ we substract the series for $e^{-x}$ from $e^x$ and divide the result by 2

Thus, $\begin{array}{rcl}{e}^x-e^{-x} & =& \sum _{k=0}^{\infty }\frac{1}{k!}{x}^k- \sum _{k=0}^{\infty }\frac{1}{k!}{(-x)}^k \\ & =& \sum _{k=0}^{\infty }\frac{1}{k!}[x^k-(-x^k\left)\right]\\ & =& 2x+\frac{2x^3}{3!}+\frac{2x^5}{5!}+\frac{2x^7}{7!}+...\end{array}$

Then dividing by $2$ we get

$\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+...$ 

implies that,  $\frac{e^x-e^{-x}}{2}=\sum _{k=0}^{\infty }\frac{1}{(2k+1)!}{x}^{2k+1}$